DF and AC are similar
and
BC and EF are similar sides
So make a proportion
\(\sf{ \frac{AC}{DF} = \frac{BC}{EF} }\)
Plug in the numbers
\(\sf{ \frac{9}{DF} = \frac{15}{25} }\)
Cross multiply
\(\sf{DF \times 15 = 9 \times 25}\)
Isolate DF and simplify it
\(\sf{DF = 15}\)
And so your final answer is
\(\boxed{\bf{15 ~centimeters}}\)
